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Parametric
Equations

Master how to define curves using a parameter \( t \), and learn to find derivatives \( \frac{dy}{dx} \) along parametric paths.

AP Calculus BC

What Are Parametric Equations?

Instead of writing \( y \) as a function of \( x \), we express both coordinates as functions of a third variable called a parameter, typically \( t \):

As \( t \) varies, the point \( (x, y) \) traces out a curve in the plane. The parameter \( t \) often represents time, but it can be any independent variable.

Why Use Parametric Form?

Curves that fail the VLT
Circles, ellipses, and figure-eights are not functions of \( x \). Parametric form handles them naturally.
Direction and speed
The parameter \( t \) encodes how fast and in which direction the point moves along the curve.
Physics modeling
Projectile paths, planetary orbits, and particle motion are naturally described by time-dependent coordinates.

Tracing a Parametric Curve

Consider the parametric equations \( x = \cos(t) \), \( y = \sin(t) \) for \( t \in [0, 2\pi] \). As \( t \) increases, the point traces a unit circle counterclockwise.

\(x\) \(y\) 1 -1 1 -1 \( t=0 \) \( t=\frac{\pi}{2} \) \( t=\pi \) \( t=\frac{3\pi}{2} \)

Eliminating the Parameter

Sometimes we can eliminate \( t \) to find a Cartesian equation. For the circle example:

This confirms the curve is a circle. However, the Cartesian form loses information about direction and speed. That is why parametric differentiation is powerful: it works directly with the parameter.

The Parametric Derivative Formula

This comes from the chain rule. Since \( y \) depends on \( t \) and \( t \) determines \( x \), we have \( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy/dt}{dx/dt} \).

The Second Derivative

To find concavity, we need \( \frac{d^2y}{dx^2} \). This is not simply \( \frac{d^2y/dt^2}{d^2x/dt^2} \). Instead:

1
Compute \( \frac{dy}{dx} \) as a function of \( t \).
2
Differentiate that result with respect to \( t \).
3
Divide by \( \frac{dx}{dt} \) again.

Parametric vs. Cartesian Derivatives

Parametric Cartesian
Given \( x = f(t),\; y = g(t) \) \( y = h(x) \)
First derivative \( \frac{dy/dt}{dx/dt} \) \( \frac{dy}{dx} = h'(x) \)
Second derivative \( \frac{d}{dt}(dy/dx) \div \frac{dx}{dt} \) \( h''(x) \)
Handles loops? Yes No
Encodes direction? Yes No

Step-by-Step Process

1
Identify the parametric equations \( x = f(t) \) and \( y = g(t) \).
2
Differentiate each with respect to \( t \): find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
3
Form the ratio: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
4
Evaluate at the given value of \( t \) (or solve for specific slope conditions).
5
For the second derivative, differentiate \( \frac{dy}{dx} \) w.r.t. \( t \), then divide by \( \frac{dx}{dt} \).

Example 1: Linear Parameter

Find \( \frac{dy}{dx} \) for the curve \( x = 3t + 1 \), \( y = t^2 - 4t \).

1
\( \frac{dx}{dt} = 3 \)
2
\( \frac{dy}{dt} = 2t - 4 \)
3
\( \frac{dy}{dx} = \frac{2t - 4}{3} \)

Example 1: Visualized

\(x\) \(y\) 2 5 8 11 2 -2 -4 (7, -4) slope = 0 \(t = 0\)

Example 2: Trigonometric Parameter

Find \( \frac{dy}{dx} \) for the ellipse \( x = 3\cos(t) \), \( y = 2\sin(t) \).

1
\( \frac{dx}{dt} = -3\sin(t) \)
2
\( \frac{dy}{dt} = 2\cos(t) \)
3
\( \frac{dy}{dx} = \frac{2\cos(t)}{-3\sin(t)} = -\frac{2}{3}\cot(t) \)

Example 3: Second Derivative

Given \( x = t^2 \), \( y = t^3 - 3t \), find \( \frac{d^2y}{dx^2} \).

1
First derivative: \( \frac{dy}{dt} = 3t^2 - 3 \), \( \frac{dx}{dt} = 2t \), so \( \frac{dy}{dx} = \frac{3t^2 - 3}{2t} \).
2
Differentiate \( \frac{dy}{dx} \) w.r.t. \( t \): use quotient rule on \( \frac{3t^2 - 3}{2t} \).
\( \frac{d}{dt}\!\left(\frac{3t^2-3}{2t}\right) = \frac{6t(2t) - (3t^2-3)(2)}{4t^2} = \frac{12t^2 - 6t^2 + 6}{4t^2} = \frac{6t^2 + 6}{4t^2} = \frac{3(t^2+1)}{2t^2} \)
3
Divide by \( \frac{dx}{dt} = 2t \): \( \frac{d^2y}{dx^2} = \frac{3(t^2+1)}{2t^2} \cdot \frac{1}{2t} = \) \( \frac{3(t^2+1)}{4t^3} \)

Tangent Lines to Parametric Curves

Once you have \( \frac{dy}{dx} \) at a specific \( t = t_0 \), the tangent line at the point \( (f(t_0), g(t_0)) \) is:

Horizontal tangent
\( \frac{dy}{dt} = 0 \) and \( \frac{dx}{dt} \neq 0 \)
Vertical tangent
\( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} \neq 0 \)

Common Mistakes to Avoid

Wrong second derivative
Writing \( \frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2} \). You must differentiate \( \frac{dy}{dx} \) w.r.t. \( t \) and divide by \( \frac{dx}{dt} \).
Forgetting the denominator
Computing \( \frac{dy}{dt} \) alone without dividing by \( \frac{dx}{dt} \). Always form the ratio for \( \frac{dy}{dx} \).
Ignoring \( dx/dt = 0 \)
When \( \frac{dx}{dt} = 0 \), the slope is undefined: there may be a vertical tangent or cusp, not a zero slope.

Summary of Key Formulas

Quantity Formula Condition
First derivative \( \frac{dy/dt}{dx/dt} \) \( dx/dt \neq 0 \)
Second derivative \( \frac{\frac{d}{dt}(dy/dx)}{dx/dt} \) \( dx/dt \neq 0 \)
Horizontal tangent \( dy/dt = 0 \) \( dx/dt \neq 0 \)
Vertical tangent \( dx/dt = 0 \) \( dy/dt \neq 0 \)
Tangent line \( y - y_0 = m(x - x_0) \) Evaluate at \( t = t_0 \)

Question 1 of 5

If \( x = t^2 + 1 \) and \( y = 3t - 5 \), what is \( \frac{dy}{dx} \)?

Question 2 of 5

For the curve \( x = \sin(t) \), \( y = \cos(t) \), what is the slope \( \frac{dy}{dx} \) at \( t = \frac{\pi}{6} \)?

Question 3 of 5

Given \( x = e^t \) and \( y = e^{2t} \), what is \( \frac{d^2y}{dx^2} \)?

Question 4 of 5

For the curve \( x = t^3 \), \( y = t^2 \), at what value of \( t \) does a horizontal tangent occur?

Question 5 of 5

The parametric equations \( x = 4\cos(t) \), \( y = 3\sin(t) \) describe an ellipse. What is the equation of the tangent line at \( t = \frac{\pi}{2} \)?